1) sin^6(x) + cos^6(x) = sin^2(x)cos^2(x)
(sin^2(x) + cos^2(x))(sin^4(x) - sin^2(x)cos^2(x) + cos^4(x)) = sin^2(x)cos^2(x)
Using the trigonometric identity sin^2(x) + cos^2(x) = 1, we get:
1 * (sin^4(x) - sin^2(x)cos^2(x) + cos^4(x)) = sin^2(x)cos^2(x)
sin^4(x) - sin^2(x)cos^2(x) + cos^4(x) = sin^2(x)cos^2(x)
sin^4(x) + cos^4(x) = 2sin^2(x)cos^2(x)
(sin^2(x) + cos^2(x))^2 - 2sin^2(x)cos^2(x) = 2sin^2(x)cos^2(x)
1 - 2sin^2(x)cos^2(x) = 2sin^2(x)cos^2(x)
2sin^2(x)cos^2(x) + 2sin^2(x)cos^2(x) = 1
4sin^2(x)cos^2(x) = 1
sin^2(x)cos^2(x) = 1/4

2) sin(x) + cos(x) = 0.3
Square both sides:
(sin(x) + cos(x))^2 = 0.3^2
sin^2(x) + 2sin(x)cos(x) + cos^2(x) = 0.09
1 + 2sin(x)cos(x) = 0.09
2sin(x)cos(x) = 0.09 - 1
2sin(x)cos(x) = -0.91
sin(2x) = -0.91
2x = arcsin(-0.91)
x ≈ -1.11 radians

3) tg(x) + ctg(x) + tg^2(x) + ctg^2(x) + tg^3(x) + ctg^3(x) = 6
tg(x) + ctg(x) = tg(x) + 1/tg(x) = 2
tg^2(x) + ctg^2(x) = (tg(x))^2 + (1/tg(x))^2 = tg^2(x) + 1 + cot^2(x) = tg^2(x) + ctg^2(x) = 2^2 - 2 = 2
tg^3(x) + ctg^3(x) = tg^3(x) + 1/tg^3(x). Let's denote tg(x) = y. So y + 1/y = 2.
Now solve the quadratic equation y^2 - 2y + 1 = 0, and get possible values for y.
tg(x) = 1 and ctg(x) = 1, therefore tg^3(x) + ctg^3(x) = 2
Summing up all the terms gives:
2 + 2 + 2 = 6

Therefore, the solutions are x ≈ -1.11 radians and x = 1.